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I don't understand what's wrong in the following argument.

Assume that we have a no-arbitrage market where the following products are traded:

  • a risky asset $S$,
  • a risk-free bond $B$,
  • an American put option $P$ with finite maturity $T$ and payoff $K$. Its underlying is $S$.

Now, according to the first fundamental theorem of asset pricing, there exists an equivalent probability measure $\mathbb Q$, under which the two-dimensional process $(\frac{S}{B},\frac{P}{B})$ is a martingale. But then, $P$ would have the same fair price as an European put option, and this is (as far as I know) false.

Where is the mistake? Have I misunderstood the statement of the theorem?

EDIT: Perhaps the key point is that when we price an American option, we do not assume that, when exercised early, another one is available (and buyable) in the market. In my argument, on the contrary, we do assume that. Does this make some sense?

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  • $\begingroup$ Hmm. Usually I see the FTAP applied to European derivatives, I am not sure how to apply it to path dependent payoffs... Anybody know? $\endgroup$ – noob2 Sep 15 at 10:51
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European Contracts

It's a really important question and as @noob2 commented, the FTAP is normally applied to European-style derivatives, even if they are (strongly) path-dependent, including barrier options and Asian options. The idea is always the same, $V_t=B_t\mathbb{E}^\mathbb{Q}\left[\frac{\xi_T}{B_T}\Big|\mathcal{F}_t\right]$, that is the derivative's price process is the conditional risk-neutral expectation of the future discounted payoff, $\xi_T$ (which may depend on hitting barrier levels etc). It essentially follows from the fact that for any integrable random variable $X$, the process $\mathbb{E}[X|\mathcal{F}_t]$ is a martingale. If you add the price process $V_t$ to a market where discounted asset prices are martingales, then you don't introduce new arbitrage (by the FTAP) and thus, $V_t$ is a no-arbitrage compatible price for trading the payoff $\xi$. So why is the early exercise such a problem for the martingale property?

Buyer's Price for American contracts

The price of a payoff that can be exercised at any time is much more related to something like $$U_t=\sup_{\tau\in \mathcal S_{t,T}}\left\{\mathbb{E}^\mathbb{Q}\left[\frac{\xi_\tau}{B_\tau}\bigg|\mathcal{F}_t\right]\right\},$$ where the supremum is taken over the set of all stopping times (exercise strategies) with values in $\{t,...,T\}$. Of course, $U_T=\frac{\xi_T}{B_T}$. This process $U$ relates to Snell's Envelope. A stopping time $\tau$ is optimal if $U_t=\mathbb{E}^\mathbb{Q}\left[\frac{\xi_\tau}{B_\tau}\Big|\mathcal{F}_t\right]$. The option price would be $B_tU_t$.

Two important properties:

  • $U$ dominates the payoff $\xi$: we know this, an American option is always worth at least its immediate payoff (by no-arbitrage)
  • $U$ is a supermartingale: that causes the problem with the FTAP (see below)

Optimal Exercise

Let's (recursively) define the following stopping time, $\tau_t^*$ via $\tau_T^*=T$ and for $t<T$ as \begin{align*} \tau^*_t=\begin{cases} t & \text{if } \frac{\xi_t}{B_t}\geq \mathbb{E}^\mathbb{Q}\left[\frac{\xi_{\tau_{t+1}^*}}{B_{\tau_{t+1}^*}}\bigg|\mathcal F_t\right], \\\\ \tau_{t+1}^* & \text{if }\frac{\xi_t}{B_t}< \mathbb{E}^\mathbb{Q}\left[\frac{\xi_{\tau_{t+1}^*}}{B_{\tau_{t+1}^*}}\bigg|\mathcal F_t\right]. \end{cases} \end{align*} So what does $\tau^*_t$ mean economically? If the immediate payoff $\xi_t$ is bigger than the continuation value, $B_t\mathbb{E}^\mathbb{Q}\left[\frac{\xi_{\tau_{t+1}^*}}{B_{\tau_{t+1}^*}}\bigg|\mathcal F_t\right]$, then exercise the option ($\tau_t^*=t$) and otherwise, keep holding the option.

Two properties related to this stopping time

  • $U_t=\mathbb{E}^\mathbb{Q}\left[\frac{\xi_{\tau_t^*}}{B_{\tau_t^*}}\bigg|\mathcal{F}_t\right]$, i.e. $\tau_t^*$ is optimal
  • $U_t=\max\{\frac{\xi_t}{B_t},\mathbb{E}^\mathbb{Q}[U_{t+1}|\mathcal{F}_t]\}$ starting with $U_T=\frac{\xi_T}{B_T}$. This property is also used to define Snell's envelope and captures the entire idea of binomial trees: start at maturity and work backwards, comparing every time whether exercise is optimal (the payoff $\frac{\xi_t}{B_t}$ is larger) or the continuation value of keeping the option for another period. This representation also immediately tells you that $U$ is a supermartingale: $$U_t=\max\left\{\frac{\xi_t}{B_t},\mathbb{E}^\mathbb{Q}[U_{t+1}|\mathcal{F}_t]\right\}\geq \mathbb{E}^\mathbb{Q}[U_{t+1}|\mathcal{F}_t]$$

Summary

Because you can exercise at any time, your option value is a supremum over all exercise strategies (stopping times). The FTAP and martingale pricing would simply take the payoff and construct the corresponding price process by discounting and conditioning but for American options you have to think about the optimal stopping time.

A few notes

  • The notes above are kind of from a buyer's perspective. You can take a hedger's perspective and show that a seller has the same price if the buyer behaves optimally.
  • As always, if markets are incomplete, $\mathbb Q$ is not unique and infinitely many fair prices may exist.
  • All the statements above are proven via backward induction: show that it holds for $t=T$ (normally trivially by construction) and show that if it holds for $t+1$, then it also holds for $t$.
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  • $\begingroup$ Thank you. Your answer is very clear, but there is still something I don't get. What happens when an American put is added to the market (buyable at any time)? If the no-arbitrage assumption is still true, the put's price should be at the same time a martingale (by FTAP) and a strict supermartingale (by your argument). The only conclusion I can draw is that there is no such market. Or, I have misunderstood the statement of FTAP, and it is not applicable in this case, but why? $\endgroup$ – chalk Sep 16 at 8:33
  • $\begingroup$ @chalk You raise a good point. The trick is the hedger's position which I kind of brushed over. So, the buyer found his optimal exercise strategy, $\tau_t^*$. Now, we can show that you as a seller can hedge the American option (if the buyer behaves optimally) without creating arbitrage. Thus, both prices (buyer's and seller's) agree and, importantly, are free of arbitrage. The proof applies Doob's decomposition to $U_t$. $\endgroup$ – Kevin Sep 16 at 9:37

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